Please help with these 3! Will mark brainliest. Thank you! How is the graph of y equals -8x^2 - 2. different from the graph of y equals negative 8x squared.? (1 point) It is shifted 2 units to the leftIt is shifted 2 units to the right. *** It is shifted 2 units up. It is shifted 2 units down. A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y equals negative 0.06 x squared plus 9.6 x plus 5.4 where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land? Round your answer to the nearest hundredth.4.30 m160.56 m160.23 m13.94 m A physics student stands at the top of hill that has an elevation of 37 meters. He throws a rock and it goes up into the air and then falls back past him and lands on the ground below. The path of the rock can be modeled by the equation y equals negative 0.02 x squared plus 0.8 x plus 37 where x is the horizontal distance, in meters, from the starting point on the top of the hill and y is the height, in meters, of the rock above the ground. How far horizontally from its starting point will the rock land? Round your answer to the nearest hundredth. 37.00 m 67.43 m27.43 m37.78 m

#1
The two functions are: [tex]y=-8 x^{2} -2[/tex] and [tex]y=-8 x^{2} [/tex] so the difference is that -2. Adding or subtracting a constant moves a function up (if the constant is positive) or down (if the constant is negative). So adding -2 means the function will shift down 2 units. That is, the last choice given.

#2
The horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation: [tex]y=-.06 x^{2} +9.6x+5.4[/tex]. To solve this we can use the quadratic formula.

The quadratic formula is: [tex]x={ \frac{-b(plus minus) \sqrt {b^{2} -4ac}}{2a} } [/tex]. We need to determine the values of a, b and c from the equation we are trying to solve. a is the coefficient (number in front of) [tex] x^{2} [/tex], b is the coefficient of x and c is the constant (the number by itself). So in this problem we have:
a=-.06
b=9.6
c=5.4
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
[tex]x={ \frac{-9.6(plus minus) \sqrt {9.6^{2} -4(-.06)(5.4)}}{2(-.06)} }[/tex]
[tex]x= \frac{-9.6plusminus \sqrt{92.16+1.296} }{-.12} [/tex]
[tex]x= \frac{-9.6plusminus 9.6672}{-.12}[/tex]
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
(-9.6-9.6672643)/-.12=160.56

c)
This part is done the same way as part b. Again, the horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation: [tex]y=-.02 x^{2} +.8x+37[/tex]. To solve this we can use the quadratic formula.

The quadratic formula is: [tex]x={ \frac{-b(plus minus) \sqrt {b^{2} -4ac}}{2a} } [/tex]. We need to determine the values of a, b and c from the equation we are trying to solve. a is the coefficient (number in front of) [tex] x^{2} [/tex], b is the coefficient of x and c is the constant (the number by itself). So in this problem we have:
a=-.02
b=.8
c=37
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
[tex]x={ \frac{-.8(plus minus) \sqrt {.8^{2} -4(-.02)(37)}}{2(-.02)} }[/tex]
[tex]x= \frac{-.8plusminus \sqrt{.64+.2.96} }{-.04} [/tex]
[tex]x= \frac{-.8plusminus 1.8973}{-.04}[/tex]
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
(-.8-1.8973)/-.04=67.43